$$\newcommand\cx{[\mathbf{A}^{op}, \mathbf{B}^{op}]}$$ $$\newcommand\cy{[\mathbf{A}, \mathbf{B}]^{op}}$$

Theorem: For categories $\mathbf{A}$ and $\mathbf{B}$, $\cx$ is isomorphic to $\cy$.

Proof: We'll first need to define a functor $\mathcal{F} : \cx \rightarrow \cy$. Let $F$ be an object of $\cx$ -- that is a functor $\mathbf{A}^{op} \rightarrow \mathbf{B}^{op}$. Then, let $\mathcal{F}(F)$ be a functor $\mathbf{A} \rightarrow \mathbf{B}$ such that for an object $A \in \mathbf{A}$, $\mathcal{F}(F)(A) = F(A)$, and for an arrow $f$ of $\mathbf{A}$, $\mathcal{F}(F)(f) = F(f^{op})^{op}$ (flip the arrow, send it through $F$, and flip it again).

Now, having described what $\mathcal{F}$ does to objects in $\cx$ (functors), we need to describe what $\mathcal{F}$ does to arrows (natural transformations). If $\alpha : F \rightarrow G$ is a natural transformation between $F, G : \mathbf{A}^{op} \rightarrow \mathbf{B}^{op}$, We need $\mathcal{F}(\alpha)$ to be natural transformation $\mathcal{F}(G) \rightarrow \mathcal{F}(F)$ between $\mathcal{F}(G), \mathcal{F}(F) : \mathbf{A} \rightarrow \mathbf{B}$. For $A \in \mathbf{A}$, notice that $\alpha_A$ is an arrow in $\mathbf{B}^{op}$ whose domain is $F(A)$ and whose codomain is $G(A)$. So we can just define $\mathcal{F}(\alpha)_A = (\alpha_A)^{op}$, which is an arrow in $\mathbf{B}$ whose domain is $G(A) = \mathcal{F}(G)(A)$ and whose codomain is $F(A) = \mathcal{F}(F)(A)$. Then, we have a well-defined map $\mathcal{F}(\alpha)_A : \mathcal{F}(G)(A) \rightarrow \mathcal{F}(F)(A)$ for each $A$, and $\mathcal{F}(\alpha)$ is a natural transformation as prescribed.

We quickly verify that $\mathcal{F}$ is indeed a functor. If $1_F$ is the identity natural transformation on $F \in \cx$, then

$$\mathcal{F}(1_F)_A = ((1_F)_A)^{op} = (1_{F(A)})^{op} = 1_{F(A)} = 1_{\mathcal{F}(F)(A)} = (1_{\mathcal{F}(F)})_A$$

so $\mathcal{F}(1_F) = 1_{\mathcal{F}(F)}$.

Suppose $F \xrightarrow{\alpha} G \xrightarrow{\beta} H$ are natural transformations. Then we have

$$\mathcal{F}(\beta \alpha)_A = ((\beta \alpha)_A)^{op} = (\beta_A \alpha_A)^{op} = (\alpha_A)^{op} (\beta_A)^{op} = \mathcal{F}(\alpha)_A \mathcal{F}(\beta)_A$$

(composed in the opposite order because $\mathcal{F}$ is contravariant).

One can obtain the inverse functor by applying the identical construction starting with $\mathcal{A}^{op}$ and $\mathcal{B}^{op}$. In this way, we get a functor $\mathcal{G} : [\mathbf{A}, \mathbf{B}] \rightarrow [\mathbf{A}^{op}, \mathbf{B}^{op}]^{op}$. But, this can equally well be described as a functor $\mathcal{G} : \cy \rightarrow \cx$. Either way, $\mathcal{G}$ is defined by identical rules to $\mathcal{F}$. If $F \in \cx$, recall $\mathcal{F}(F)(A) = F(A)$ and $\mathcal{F}(F)(f) = F(f^{op})^{op}$. So, $\mathcal{G}(\mathcal{F}(F))(A) = \mathcal{F}(F)(A) = F(A)$ and $\mathcal{G}(\mathcal{F}(F))(f) = (\mathcal{F}(F)(f^{op}))^{op} = (F(f)^{op})^{op} = F(f)$. So, $(\mathcal{G} \circ \mathcal{F})(F) = F$. If $\alpha : F \rightarrow G$ is a natural transformation, recall that we have $\mathcal{F}(\alpha)_A = (\alpha_A)^{op}$. So, $\mathcal{G}(\mathcal{F}(\alpha))_A = (\mathcal{F}(\alpha)_A)^{op} = ((\alpha_A)^{op})^{op} = \alpha_A$. So $(\mathcal{G} \circ \mathcal{F})(\alpha) = \alpha$. Together, this means that $\mathcal{G} \circ \mathcal{F}$ is the identity functor on $\cx$. It follows symmetrically that $\mathcal{F} \circ \mathcal{G}$ is the identity on $\cy$. Thus, $\cx$ and $\cy$ are isomorphic.